$\sum\limits_{n=1}^{\infty } \dfrac{n(x-1)^n}{(n+3)\cdot(-3)^n} $ What is the interval of convergence of the series? Choose 1 answer: Choose 1 answer: (Choice A) A $-2<x<4$ (Choice B) B $-6<x<0$ (Choice C) C $-6 \le x \le 0$ (Choice D) D $-2 \le x \le 4$
Explanation: We use the ratio test. For $x\neq 1$, let $a_n= \dfrac{n(x-1)^n}{(n+3)\cdot(-3)^n} $. $\lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|= \left| \dfrac{x-1}{3}\right| $ The series converges when $\left|\dfrac{x-1}{3}\right| <1$, which is when $-2<x<4$. Now let's check the endpoints, $x=-2$ and $x=4$. Letting $x=-2$, we get the series: $\begin{aligned} \sum\limits_{n=1}^{\infty }\dfrac{n(-2-1)^n}{(n+3)\cdot(-3)^n} &=\sum\limits_{n=1}^{\infty } \dfrac{n(-3)^n}{(n+3)\cdot(-3)^n} \\\\ &=\sum\limits_{n=1}^{\infty } \dfrac{n}{(n+3)} \end{aligned}$ By the $n^{\text{th}}$ -term test, we know this series diverges. Letting $x=4$, we get the series: $\begin{aligned} \sum\limits_{n=1}^{\infty }\dfrac{n(4-1)^n}{(n+3)\cdot(-3)^n} &=\sum\limits_{n=1}^{\infty } \dfrac{n(3)^n}{(n+3)\cdot(-3)^n} \\\\ &=\sum\limits_{n=1}^{\infty } \dfrac{(-1)^n\cdot n(-3)^n}{(n+3)\cdot(-3)^n} \\\\ &=\sum\limits_{n=1}^{\infty } \dfrac{(-1)^n\cdot n}{(n+3)} \end{aligned}$ By the $n^{\text{th}}$ -term test, we know this series diverges. In conclusion, the interval of convergence is $-2<x<4$.